The homogeneous Bessel differential equation

$$x^2 f''(x) + xf'(x) + (x^2 - \nu^2) f(x) = 0$$

does not change if $x$ is substituted with $-x$. So, it could be expected that it is the same as regards the solutions: $J_{\nu} (-x) = J_{\nu}(x)$. However, it is apparently not the case!

The Bessel function of the first kind $J_{\nu} (x)$ is even or odd respectively when $\nu$ is integer and even or odd. It is the same for: the modified Bessel function of the first kind $I_{\nu} (x)$ (which is proportional to $J_{\nu} (x)$), the Bessel function of the second kind $Y_{\nu} (x)$ (which depends on $J_{\nu} (x)$ and $J_{-\nu} (x)$) and the modified Bessel function of the second kind $K_{\nu} (x)$ (which depends on $I_{\nu} (x)$ and $I_{-\nu} (x)$).

When $\nu$ is complex, instead:

$$J_{\nu}(-x) = e^{j \pi \nu} J_{\nu}(x)$$

as well as

$$I_{\nu}(-x) = e^{j \pi \nu} I_{\nu}(x)$$

$e^{j \pi \nu}$ are in general complex values.

How is it possible that $J_{\nu}(-x) \neq J_{\nu}(x)$ as well as for the other three functions? If the equation does not change when $x \rightarrow -x$, how can instead the solutions change?

From the symmetry you only know that when $f(x)$ is a solution, then also $f(-x)$ is a solution. Then also the linear combinations $f(x)+f(-x)$ and $f(x)-f(-x)$ are solutions because of the linearity of the ODE, which are, if not zero, even and odd functions respectively. As odd and even are obviously linearly independent, one can try to get an even-odd pair of basis solutions.

With the Bessel equation you have the additional difficulty of the singularity at zero. Thus you get the additional distinction of functions that can be continued to zero and functions that diverge at zero. Of the even-odd basis pair, at most one can be continued to zero, which one depends on the nature of the equation.

  • "From the symmetry you only know that when $f(x)$ is a solution, then also $f(−x)$ is a solution". Ok! But I can not figure out: why? – BowPark Nov 9 at 9:41
  • To cite your question: "does not change if $x$ is substituted with $−x$", just set $g(x)=f(-x)$ to get $g'(x)=-f'(-x)$, $g''(x)=f''(-x)$ and if you insert this into the equation for $g$, you get $$(-x)^2f''(-x)+(-x)f'(-x)+((-x)^2-ν^2)f(-x)=0$$ which is true if $f$ is a solution. – LutzL Nov 9 at 9:49

Your Answer

 

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.